One thousand independent rolls of a fair die will be made. compute an approximation to the probability that the number 6 will appear between 150 and 200 times inclusively. if the number 6 appears exactly 200 times, find the probability that the number 5 will appear less than 150 times.

The probability that the number 6 will appear between 150 and 200 times inclusively is 0.9258

The probability that the number 5 will appear less than 150 times is 0.1762.

Solution:

Given, One thousand independent rolls of a fair die will be made.

First, let’s compute an approximation to the probability that number 6 will appear between 150 and 200 times.  

The die is rolled 1,000 times. These are independent trials which are Bernoulli trials since a die shows either a 6 or not.  

Thus, n = 1000. The probability of success is constant, p = 1/6. We can use the binomial model.  

Let X be the number of times the die shows a 6.  

We can use the normal approximation to the binomial model  

Because np = 1000 • 1/6 > 10 and n(1 − p) = 1000 • 5 6 > 10. x ∼ Bin(166.7, 11.79).

But, using the normal approximation, let Y be normally distributed using the parameters from the binomial model.

That is, in the normal model

So, let y ∼ N(166.7, 11.79).

There is one small complication. We are using a continuous distribution to model a discrete distribution. We must correct for this discrepancy using the continuity correction which is bolded in the equation below.  

P(150 ≤ Y ≤ 200) = P(Y ≤ 200) − P(Y ≤ 150)  

Now, we consider the probability that we obtain less than 150 6s if exactly 200 5s have been shown.

This is a conditional probability problem. Let Y be the number of 6s that appear and let X be the number of 5s that appear.  

Then we need to compute P(Y < 150|X = 200).  

That is, we consider that 200 of the rolls show a 5, therefore, we have 800 trials remaining that can display anything except a 5, so n = 800 now.  

Since we have already considered those 200 trials where a 5 appears, the probability of success (a 6 appears) is p = 1/5.  

Again, we use the normal approximation due to the large value for n and since the success/failure conditions hold (np > 10 and n(1 − p) > 10).  

Let Y ∼ N(160, 11.31).

Hence, the probabilities are 0.9258 and 0.1762.

stops on an even number.